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Does anybody know an easy explanation of the proof of Artin's vanishing theorem (that the etale cohomology of an affine variety of dimension $n$ over an algebraically closed field vanishes in degrees $>n$, or of any other version of this statement)? I have found some proofs; all of them are step by step, and it is not clear to me which of these steps are the most important ones. So, what is the central idea here?

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I was curious myself after learning this result sometime ago from Lazarsfeld's book on positivity (he calls it the Artin-Grothendieck theorem). The corresponding statement for smooth varieties over the complex numbers and singular cohomology (theorem of Andreotti-Frankel) follows from the fact that Morse theory shows that the variety is homotopy equivalent to a CW-complex with no cells in dimensions $>n$.

The etale cohomology counterpart works more generally for constructible sheaves. This is probably not very helpful, but here is a sketch of the argument from Lazarsfeld (he uses constructible sheaves in the complex topology, but it should adapt to etale sheaves):

  1. Reduce to the affine space by choosing a finite map $X\to \mathbb{A}^n$ using Noether normalization (already here it is crucial to work with constructible sheaves, not constant sheaves, so we clearly gain something from generalization),
  2. Prove the result for $\mathbb{A}^1$,
  3. Prove the result for $\mathbb{A}^n$ by induction on $n$ using the Leray spectral sequence. Here the crucial observation is that if we choose a sufficiently generic linear projection $\mathbb{A}^n\to \mathbb{A}^{n-1}$, then the stalks of the higher direct images will compute cohomology on the fibers.

So all in all it is a typical example of devissage, which I usually to dislike but slowly learn to appreciate. I think from the outline it is clear which are the key ideas, but I would still really like to see a conceptual proof.

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Thank you! Part 3 of this argument seems to be the most difficult one; I should think about it. – Mikhail Bondarko Oct 8 '12 at 9:11
Devissage is a bit like induction, which can also be confusing. Most of the work goes into the induction step (step 3). – Donu Arapura Oct 8 '12 at 11:12
The other thing that I should point out is that for this devissage argument to work, one has to prove it for general coefficients. – Donu Arapura Oct 8 '12 at 11:42

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